Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
  public static void main(String args[]) {
    Scanner scan = new Scanner(System.in);
    double[] P = new double[scan.nextInt()];
    for (int i = 0; i < P.length; i++)
      P[i] = scan.nextDouble();
    scan.close();

    System.out.println(headsOverTailsProbability(P));
  }

  private static double headsOverTailsProbability(double[] P) {
    if (P.length == 0) return 0;

    int N = P.length;
    double[][] dp = new double[N + 1][N + 1];
    dp[0][0] = 1;
    for (int i = 1; i <= N; i++) {
      double headProbability = P[i - 1];
      double tailProbability = 1 - headProbability;
      dp[i][0] = tailProbability * dp[i-1][0];
      for (int j = 1; j <= N; j++)
        dp[i][j] = dp[i-1][j-1] * headProbability + dp[i-1][j] * tailProbability;
    }

    double result = 0;
    for (int i = N/2 + 1; i <= N; i++)
      result += dp[N][i];

    return result;
  }
}